CPM Homework Help : APCALC Problem 2-125

Determine if the following functions are even, odd, or neither. Explain how you determined your choice.

Hint:

Even and Odd Functions

A function $f$ is an even function if, for all $x$ in its domain, $f\left(–x\right) = f\left(x\right)$ .

A function $f$ is an odd function if, for all $x$ in its domain, $f\left(–x\right) = –f\left(x\right)$ .

Example: If $f\left(x\right)=2x^3+\sin\left(x\right)$ , then $f\left(-x\right)=2\left(-x\right)^3+\sin\left(-x\right)$

$=2\left(-1\right)^3x^3+\left(-1\right)\sin\left(x\right)$

$=-2x^3-\sin\left(x\right)=-\left(2x^3+\sin\left(x\right)\right)$

Therefore $f\left(–x\right) = –f\left(x\right)$ , so $f$ is odd.

Step 1(a):

$y=\sin^2x=(\sin{x})(\sin{x})$

Hint (a):

$y=\sin x$ is an odd function. Definition of odd functions: $f(−x) = −f(x)$ Therefore, $\sin(−x) = −{\sin}(x)$ .

Step 2(a):

Explore what $y = \sin^2(−x)$ looks like:
$y=\sin^2(−x)$
$=(\sin(−x))(\sin(−x))$
$=(−\sin{x})(−\sin{x})$
$=\sin^2x$

Answer (a):

Hence $\sin^2(−x)=\sin^2(x)$ .
This is the definition of an even function: $f(−x) = f(x)$ .

Step 1(b):

Test for even: $f(a) = f(−a)$

$f(a)=\frac{a^{2}+1}{a^{3}-2a}$

$f(-a)=\frac{(-a)^{2}+1}{(-a)^{3}-2(-a)}=\frac{a^{2}+1}{-a^{3}+2a}\ ,$

$f(a)\ne f(−a)$ ; not even

Step 2(b):

Test for odd: $f(−a) = −f(a)$ ...